Question

Pressure ${\text{P}}_1$ is applied on a sphere of radius $5\text{cm}$. Keeping the thrust constant, if the radius of the sphere is decreased to $2.5\text{cm}$, the new pressure is found out to be ${\text{P}}_2$. Find ${\text{P}}_2:{\text{P}}_1$.

• A. 4

Answer 1

The correct option is A 4
Given:
The initial radius of the sphere (${\text{r}}_1$) = $5\text{cm}$
The final radius of the sphere (${\text{r}}_2$) = $2.5\text{cm}$

Surface area of a sphere of radius r is given by:
$A=4\pi {\text{r}}^2$
$⟹A\propto {\text{r}}^2\left(\text{i}\right)$

Pressure $\left(\text{P}\right)$ on a body is given by the ratio of thrust to the area $\left(\text{A}\right)$ of a body.
$\text{P}\propto \frac{1}{\text{A}}\left(\text{ii}\right)$

Using (i) and (ii),
$\text{P}\propto \frac{1}{{\text{r}}^2}$
$⟹\frac{{\text{P}}_2}{{\text{P}}_1}=\frac{{\text{r}}_1^2}{{\text{r}}_2^2}$

Substituting the values, we get:
$\frac{{\text{P}}_2}{{\text{P}}_1}=\frac{5^2}{(2.5{)}^2}$
$\frac{{\text{P}}_2}{{\text{P}}_1}=2^2=4$